CS50 week 2 Caesar(cipher printf question)

Question

#include <stdio.h>
#include <stdlib.h>
#include <cs50.h>
#include <ctype.h>
#include <string.h>

int main( int argc,string argv[]){

    int k;

    if (argc == 2) {

        k = atoi(argv[1]);
    }
    else
    {
        printf("Usage: ./caesar k");
    }
        string p = get_string();

       printf("Plaintext: %s\n",p);

        char c[100]; 
        for(int i = 0,n = strlen(p);i < n;i++)
        {

        if(islower(p[i]) && isalpha(p[i]))
        { 
             c[i] = (p[i] - 'a'  + k) % 26 + 97;
        }    

         if(isupper(p[i]) && isalpha(p[i]))
         {
             c[i] = (p[i] - 'A' + k) % 26 + 65; 
         }
          {  
           printf("%c",c[i]); 
          }  

        }   
      }  
}  
  printf("\n");  

  return 0;
}

In my printf statement where I want to print my Ciphertext, the printf function's output is behaving different to what I have anticapated.

output without Ciphertext before %c

Answer 1

You seem to build a new, "encrypted" string from input.

You would then print that string as a whole after the loop.

An alternative is to print the ciphertext: before the loop (so it's printed once), and within the loop print the encrypted characters one at a time, later print a newline after the loop. That one would not require storing the whole ciphertext.

BTW, your processed string does not seem to be null terminated. Also, what happens if your input string is 100 characters or longer? Using an arbitrary buffer size is not a good idea, especially not if you don't enforce this maximum length (e.g. truncate the string if it's too long, or throw an error, but there's no such thing in the specification).

Answer 2

You are printing Ciphertextext: for every character you encrypt. That is, since the print statement with Ciphertextext: is in for loop, it will print Ciphertextext: for every character you encrypt.

To illustrate, consider this,

for(int i = 0; i < 5; i++)
{
    printf("i is: %i", i); 
}

Here output will be,

i is: 0i is: 1i is: 2i is: 3i is: 4 

Did you notice that there is no \n or space after %i in my print statement ?

Contrast that with this,

for(int i = 0; i < 5; i++)
{
    printf("%i", i); 
}

Here output will be,

01234

Again, notice that there is no space or \n after %i in print statement. Hence the next statement is printed on the same line right next to the previous print statement.

I hope this gives you a hint as to where you are going wrong.